Question

9. The product of two numbers is 360, If one numnber is l6 less than double the other
number, find the two possible number pairs.

Answer 1

Let the two numbers be \( x \) and \( y \). We have the following relationships:

1. The product of the two numbers:
\[
xy = 360
\]

2. One number is 16 less than double the other number. We can express this as:
\[
y = 2x - 16
\]

Now, we can substitute the expression for \( y \) into the first equation:

\[
x(2x - 16) = 360
\]

Expanding this gives:

\[
2x^2 - 16x = 360
\]

Now, rearranging brings all terms to one side:

\[
2x^2 - 16x - 360 = 0
\]

We can simplify this equation by dividing all terms by 2:

\[
x^2 - 8x - 180 = 0
\]

Next, we will use the quadratic formula, which is given by

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, \( a = 1 \), \( b = -8 \), and \( c = -180 \). First, we calculate the discriminant:

\[
b^2 - 4ac = (-8)^2 - 4(1)(-180) = 64 + 720 = 784
\]

Now we apply the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{784}}{2 \cdot 1} = \frac{8 \pm 28}{2}
\]

This gives us two possible solutions for \( x \):

1.
\[
x = \frac{36}{2} = 18
\]

2.
\[
x = \frac{-20}{2} = -10
\]

Now we will find the corresponding values of \( y \) for these \( x \) values:

1. When \( x = 18 \):
\[
y = 2(18) - 16 = 36 - 16 = 20
\]

2. When \( x = -10 \):
\[
y = 2(-10) - 16 = -20 - 16 = -36
\]

Thus, the two pairs of numbers that multiply to give 360 and satisfy the second condition are:

1. \( (18, 20) \)
2. \( (-10, -36) \)

So, the two possible number pairs are \( (18, 20) \) and \( (-10, -36) \).